I've looked around here for a while trying all different kinds of accepted answers, but they are too irrelevant for my own needs and 90% require static typing of the current date which is something I don't want to do
Context
Quite simple.
Get the current time in Spain.
Have a variable already set up which is an agreed start time.
Have a variable already set up which is an agreed end time.
If the current time in Spain is between the start-end time, proceed with saving logs
Threads checked for solution to my problem
All useless, downvoted, or irrelevant due to static typing of datetime
Python current time
Compare dates Python
Convert string into datetime
Compare date and datetime
Check if time is between two days
Convert string into datetime
Compare if datetime.timedelta is between two values
Code so far
now_time_uk = datetime.datetime.now()
current_time_spain = now_time + datetime.timedelta(hours=1)
start = datetime.time(10, 50)
end = datetime.time(16, 00)
if current_time_spain >= start or current_time_spain <= end:
print('Start logging')
else:
print('No need to log')
The code above was extracted from a thread on S.O which was an accepted answer, when I ran this, it gives a TypeError
TypeError: can't compare datetime.datetime to datetime.time
Accepted Answer Code
now_time_uk = datetime.datetime.now()
print(now_time_uk)
current_time_spain = now_time_uk + datetime.timedelta(hours=1)
start = datetime.time(10, 50)
end = datetime.time(17, 00)
print(current_time_spain.time())
if current_time_spain.time() >= start and current_time_spain.time() <= end:
print('Start logging')
else:
print('No need to log')
Like the error says you are trying to compare datetime objects with time objects which isn't possible... luckily the datetime object has a builtin method for converting to a time object, and that is .time(). So you can replace:
if current_time_spain >= start or current_time_spain <= end:
With
if current_time_spain.time() >= start or current_time_spain.time() <= end:
And it should work. Another way of doing it is setting current_time_spain from the beginning as a time object or another option is to set the start and end times to be datetime objects instead of time
You're comparing datetime object with date & time with only time , as mentioned in error
datetime.datetime.now() this returns datetime whereas datetime.time(10, 50) returns time.
If you want to only compare time then why you don't simply compare hours and minutes from now()
You can’t compare a date time within a time but you could continue to use a TimeDelta in order to create a start and end date to compare. With this solution your code could become:
now_time_uk = date time.datetime.now()
current_time_spain = now_time + datetime.timedelta(hours=1)
start_hours = 10
start_minutes = 50
end_hours = 16
end_minutes = 0
time_spain_start = current_time_spain + datetime.timedelta(hours=start_hours, minutes=start_minutes)
time_spain_end = current_time_spain + datetime.timedelta(hours=end_hours, minutes=end_minutes)
if current_time_spain >= time_spain_start or current_time_spain <= spain_time_end:
print “Start logging”
else:
print “No need to log”
In this way you can also add a variable for each parameter of the TimeDelta object in order to have different time for the end date even if the day is the next or whatever you want or need. We can imagine to have all variable days, hours, minutes, and seconds initialized to the 0 value, so you need to change only the value you need.
Python noob here
from datetime import datetime, time
now = datetime.now()
now_time = now.time()
if now_time >= time(10,30) and now_time <= time(13,30):
print "yes, within the interval"
I would like the timer to work between 10,30 AM today and 10 AM the next day. Changing time(13,30) to time(10,00) will not work, because I need to tell python 10,00 is the next day. I should use datetime function but don't know how. Any tips or examples appreciated.
The combine method on the datetime class will help you a lot, as will the timedelta class. Here's how you would use them:
from datetime import datetime, timedelta, date, time
today = date.today()
tomorrow = today + timedelta(days=1)
interval_start = datetime.combine(today, time(10,30))
interval_end = datetime.combine(tomorrow, time(10,00))
time_to_check = datetime.now() # Or any other datetime
if interval_start <= time_to_check <= interval_end:
print "Within the interval"
Notice how I did the comparison. Python lets you "nest" comparisons like that, which is usually more succinct than writing if start <= x and x <= end.
P.S. Read https://docs.python.org/2/library/datetime.html for more details about these classes.
Consider this:
from datetime import datetime, timedelta
now = datetime.now()
today_10 = now.replace(hour=10, minute=30)
tomorrow_10 = (now + timedelta(days=1)).replace(hour=10, minute=0)
if today_10 <= now <= tomorrow_10:
print "yes, within the interval"
The logic is to create 3 datetime objects: one for today 10 AM, one for right now and one for tomorrow 10 AM. Them simply checking for the condition.
An alternative to creating time objects for the sake of comparison is to simply query the hour and minute attributes:
now= datetime.now().time()
if now.hour<10 or now.hour>10 or (now.hour==10 and now.minute>30):
print('hooray')
I have two date time objects
`statrt_time` and `end_time`
my code is
if self.book_from and self.book_to:
fmt = '%Y-%m-%d %H:%M:%S'
s = datetime.strptime(self.book_from,fmt) #start date
e = datetime.strptime(self.book_to,fmt) #end date
diff = e - s
total_seconds=diff.seconds
time_diff = (total_seconds/3600.0)
no_of_units = (time_diff/4)
if(e<s):
self.units = 0
else:
self.units = math.ceil(no_of_units)
Here when I subtract time within the same day it is giving the correct difference. But when the day is changed, it is not calculating the day difference but only giving time difference. How can I add day difference also?
Use total_seconds() instead of seconds.
timedelta.seconds just shows "second" part of the difference, while total_seconds() shows the duration of the difference in seconds. See Mureinik's answer for more details.
So, use this:
total_seconds=diff.total_seconds()
total_seconds is a timedelta object which stores the difference between two datetimes using three fields - days, seconds and miliseconds. Your snippet just uses the seconds attributes instead of the entire difference. The total_seconds() method takes care of this for you and returns, well, the total number of seconds between two datatimes.
I got another way of doing.. BUT A WORK AROUND..
if self.book_from and self.book_to:
fmt = '%Y-%m-%d %H:%M:%S'
s = datetime.strptime(self.book_from,fmt) #start date
e = datetime.strptime(self.book_to,fmt) #end date
diff = e - s
days=diff.days// convert difference to day instead of seconds//
days_seconds=0
if(days>0): //Checking whether the difference exceeds a day//
days_seconds=days*24*3600 //If so convert it to seconds//
total_seconds=diff.seconds+days_seconds
time_diff = (total_seconds/3600.0)
I want to compare only time part in datetime. I have different dates with only time field to compare. Since dates are different and only time part i want to consider So i think creating two datetime object will not help.
my string as
start="22:00:00"
End="03:00:00"
Tocompare="23:30:00"
Above are strings when i convert them with datetime as
dt=datetime.strptime(start,"%H:%M:%S")
it gives
1900-01-01 22:00:00
which is default date in python.
So i need to avoid all this and want only time part. I simply need to check does my Tocompare falls between start and End
Just call the .time() method of the datetime objects to get their hours, minutes, seconds and microseconds.
dt = datetime.strptime(start,"%H:%M:%S").time()
Compare their times using datetime.time().
import datetime
start = datetime.datetime.strptime(start, '%H:%M:%S')
start = datetime.time(start.hour, start.minute,start.second)
tocompare = datetime.datetime.strptime(tocompare, '%H:%M:%S')
tocompare = datetime.time(tocompare.hour, tocompare.minute, tocompare.second)
start > tocompare # False
I have two times, a start and a stop time, in the format of 10:33:26 (HH:MM:SS). I need the difference between the two times. I've been looking through documentation for Python and searching online and I would imagine it would have something to do with the datetime and/or time modules. I can't get it to work properly and keep finding only how to do this when a date is involved.
Ultimately, I need to calculate the averages of multiple time durations. I got the time differences to work and I'm storing them in a list. I now need to calculate the average. I'm using regular expressions to parse out the original times and then doing the differences.
For the averaging, should I convert to seconds and then average?
Yes, definitely datetime is what you need here. Specifically, the datetime.strptime() method, which parses a string into a datetime object.
from datetime import datetime
s1 = '10:33:26'
s2 = '11:15:49' # for example
FMT = '%H:%M:%S'
tdelta = datetime.strptime(s2, FMT) - datetime.strptime(s1, FMT)
That gets you a timedelta object that contains the difference between the two times. You can do whatever you want with that, e.g. converting it to seconds or adding it to another datetime.
This will return a negative result if the end time is earlier than the start time, for example s1 = 12:00:00 and s2 = 05:00:00. If you want the code to assume the interval crosses midnight in this case (i.e. it should assume the end time is never earlier than the start time), you can add the following lines to the above code:
if tdelta.days < 0:
tdelta = timedelta(
days=0,
seconds=tdelta.seconds,
microseconds=tdelta.microseconds
)
(of course you need to include from datetime import timedelta somewhere). Thanks to J.F. Sebastian for pointing out this use case.
Try this -- it's efficient for timing short-term events. If something takes more than an hour, then the final display probably will want some friendly formatting.
import time
start = time.time()
time.sleep(10) # or do something more productive
done = time.time()
elapsed = done - start
print(elapsed)
The time difference is returned as the number of elapsed seconds.
Here's a solution that supports finding the difference even if the end time is less than the start time (over midnight interval) such as 23:55:00-00:25:00 (a half an hour duration):
#!/usr/bin/env python
from datetime import datetime, time as datetime_time, timedelta
def time_diff(start, end):
if isinstance(start, datetime_time): # convert to datetime
assert isinstance(end, datetime_time)
start, end = [datetime.combine(datetime.min, t) for t in [start, end]]
if start <= end: # e.g., 10:33:26-11:15:49
return end - start
else: # end < start e.g., 23:55:00-00:25:00
end += timedelta(1) # +day
assert end > start
return end - start
for time_range in ['10:33:26-11:15:49', '23:55:00-00:25:00']:
s, e = [datetime.strptime(t, '%H:%M:%S') for t in time_range.split('-')]
print(time_diff(s, e))
assert time_diff(s, e) == time_diff(s.time(), e.time())
Output
0:42:23
0:30:00
time_diff() returns a timedelta object that you can pass (as a part of the sequence) to a mean() function directly e.g.:
#!/usr/bin/env python
from datetime import timedelta
def mean(data, start=timedelta(0)):
"""Find arithmetic average."""
return sum(data, start) / len(data)
data = [timedelta(minutes=42, seconds=23), # 0:42:23
timedelta(minutes=30)] # 0:30:00
print(repr(mean(data)))
# -> datetime.timedelta(0, 2171, 500000) # days, seconds, microseconds
The mean() result is also timedelta() object that you can convert to seconds (td.total_seconds() method (since Python 2.7)), hours (td / timedelta(hours=1) (Python 3)), etc.
This site says to try:
import datetime as dt
start="09:35:23"
end="10:23:00"
start_dt = dt.datetime.strptime(start, '%H:%M:%S')
end_dt = dt.datetime.strptime(end, '%H:%M:%S')
diff = (end_dt - start_dt)
diff.seconds/60
This forum uses time.mktime()
Structure that represent time difference in Python is called timedelta. If you have start_time and end_time as datetime types you can calculate the difference using - operator like:
diff = end_time - start_time
you should do this before converting to particualr string format (eg. before start_time.strftime(...)). In case you have already string representation you need to convert it back to time/datetime by using strptime method.
I like how this guy does it — https://amalgjose.com/2015/02/19/python-code-for-calculating-the-difference-between-two-time-stamps.
Not sure if it has some cons.
But looks neat for me :)
from datetime import datetime
from dateutil.relativedelta import relativedelta
t_a = datetime.now()
t_b = datetime.now()
def diff(t_a, t_b):
t_diff = relativedelta(t_b, t_a) # later/end time comes first!
return '{h}h {m}m {s}s'.format(h=t_diff.hours, m=t_diff.minutes, s=t_diff.seconds)
Regarding to the question you still need to use datetime.strptime() as others said earlier.
Try this
import datetime
import time
start_time = datetime.datetime.now().time().strftime('%H:%M:%S')
time.sleep(5)
end_time = datetime.datetime.now().time().strftime('%H:%M:%S')
total_time=(datetime.datetime.strptime(end_time,'%H:%M:%S') - datetime.datetime.strptime(start_time,'%H:%M:%S'))
print total_time
OUTPUT :
0:00:05
import datetime as dt
from dateutil.relativedelta import relativedelta
start = "09:35:23"
end = "10:23:00"
start_dt = dt.datetime.strptime(start, "%H:%M:%S")
end_dt = dt.datetime.strptime(end, "%H:%M:%S")
timedelta_obj = relativedelta(start_dt, end_dt)
print(
timedelta_obj.years,
timedelta_obj.months,
timedelta_obj.days,
timedelta_obj.hours,
timedelta_obj.minutes,
timedelta_obj.seconds,
)
result:
0 0 0 0 -47 -37
Both time and datetime have a date component.
Normally if you are just dealing with the time part you'd supply a default date. If you are just interested in the difference and know that both times are on the same day then construct a datetime for each with the day set to today and subtract the start from the stop time to get the interval (timedelta).
Take a look at the datetime module and the timedelta objects. You should end up constructing a datetime object for the start and stop times, and when you subtract them, you get a timedelta.
you can use pendulum:
import pendulum
t1 = pendulum.parse("10:33:26")
t2 = pendulum.parse("10:43:36")
period = t2 - t1
print(period.seconds)
would output:
610
import datetime
day = int(input("day[1,2,3,..31]: "))
month = int(input("Month[1,2,3,...12]: "))
year = int(input("year[0~2020]: "))
start_date = datetime.date(year, month, day)
day = int(input("day[1,2,3,..31]: "))
month = int(input("Month[1,2,3,...12]: "))
year = int(input("year[0~2020]: "))
end_date = datetime.date(year, month, day)
time_difference = end_date - start_date
age = time_difference.days
print("Total days: " + str(age))
Concise if you are just interested in the time elapsed that is under 24 hours. You can format the output as needed in the return statement :
import datetime
def elapsed_interval(start,end):
elapsed = end - start
min,secs=divmod(elapsed.days * 86400 + elapsed.seconds, 60)
hour, minutes = divmod(min, 60)
return '%.2d:%.2d:%.2d' % (hour,minutes,secs)
if __name__ == '__main__':
time_start=datetime.datetime.now()
""" do your process """
time_end=datetime.datetime.now()
total_time=elapsed_interval(time_start,time_end)
Usually, you have more than one case to deal with and perhaps have it in a pd.DataFrame(data) format. Then:
import pandas as pd
df['duration'] = pd.to_datetime(df['stop time']) - pd.to_datetime(df['start time'])
gives you the time difference without any manual conversion.
Taken from Convert DataFrame column type from string to datetime.
If you are lazy and do not mind the overhead of pandas, then you could do this even for just one entry.
Here is the code if the string contains days also [-1 day 32:43:02]:
print(
(int(time.replace('-', '').split(' ')[0]) * 24) * 60
+ (int(time.split(' ')[-1].split(':')[0]) * 60)
+ int(time.split(' ')[-1].split(':')[1])
)